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ellipse
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HallsofIvy
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The graph of \(\displaystyle \frac{(x- x_0)^2}{a^2}+ \frac{(y- y_0)^2}{b^2}= 1\)
is an ellipse with center at \(\displaystyle (x_0, y_0)\) axis parallel to the x-axis from \(\displaystyle (x_0- a, y_0)\) to \(\displaystyle (x_0+ a, y_0)\) and axis parallel to the y-axis from \(\displaystyle x_0, y_0- a)\) to \(\displaystyle (x_0, y_0+ a)\).
is an ellipse with center at \(\displaystyle (x_0, y_0)\) axis parallel to the x-axis from \(\displaystyle (x_0- a, y_0)\) to \(\displaystyle (x_0+ a, y_0)\) and axis parallel to the y-axis from \(\displaystyle x_0, y_0- a)\) to \(\displaystyle (x_0, y_0+ a)\).
Do you not know how to do a plot or is it just this function you are unsure of?hello guys i need to plot this, but i don't know how to do it.
(x+2)^2+4(y+4)^2=1
Assuming the latter, possibly the easiest way is to re-write the equation as
(y + 4)2 = \(\displaystyle \frac{1}{4}\) ( 1 - (x+2)2 )
or
y = - 4 \(\displaystyle \pm \frac{1}{2}\sqrt{ 1 - (x+2)^2 }\)
If we note that the graph is symmetric about y = -4, we can just do the one side (say, use the + sign) and rotate the graph around the y=-4 line.
As far as the shape of the graph, we note that the form of an ellipse is
\(\displaystyle (\frac{x-x_0}{a})^2 + (\frac{y-y_0}{b})^2 = r^2\)
which is just the form of the above equation with x0 = -2, y0 = -4, a = 1, b = 1/2, and r = 1.